Its pretty simple to create a hydrogen and oxygen generator, I wonder how the diesel would like it?

 

What are you thinking? Water electrolysis?

I was wondering about the feasibility of using peroxide in a water injection type system. H2-02 gives up 02 pretty easily. Just not sure of what other effects might be....

 

Great Troy.............As if I didn't have enough ideas "floating" around in my head! Something else to experiment with this winter.........................

 

H2O2 is not free. Water plus dc is free everywhere you drive. You could even p in the tank if you had too. Salt water is even better

 

There are some interesting Hydrogen boost products out there that I have come across. I haven't tried the product yet but am currently trying to get as much practical information about topic
Hydrogen Garage Hydrogen Boost Water gives me gas

 

Never quite understood the water -> hydrogen and oxygen craze.

I look at it this way:

You are burning H2 in the engine. Generally 1/3 of your fuel is wasted as heat, 1/3 to overcome friction, and 1/3 to produce usable energy. If that entire 1/3 goes yo your alternator, they are generally quoted as about 50% efficient at making DC power. So now your DC power goes to your electrolytic cell - one of the more inefficient ways to make H2 gas - maybe 50% for a home grown system.

So for anyone keeping track of those efficiencies:

1/3 x 1/2 x 1/2 = 1/12th !!

So for every unit of hydrogen you burn, about 8 percent of that energy is available in the form of more hydrogen. Or to put it simply, if I start with a full tank of hydrogen and burn it, use the resulting power to turn the alternator and use that voltage in an electrolytic cell to make more hydrogen, then fill a second tank with the resulting hydrogen, the second tank will be only 8% as full as the first - and we aren't even talking about moving the truck yet!

 

Corey you math is incorrect for the application I posted.

1. Hydrogen can be produces with as little as a 1.5 volt battery. A 9 volt battery makes quite a bit of hydrogen with very lil amperage.

2. I can remove 1 light bulb and produce hydrogen to inject into the diesel air system without compromising the electrical system.

3. The truck along with every car on the road is making hydrogen all the time. Why do you think batteries bubble when charging or your not supposed to smoke around them and filtered water is needed to refill the batteries.

4. 1/3x1/2x1/2 is the same as .3x.5x.5= .75 That is 3/4 !

5. Where do you identify a UNIT and how do you come up with 8 percent number?

6. 1/3 wasted as heat? Hydrogen burns colder then fuel and warmer then alcohol. How can 1/3 be wasted on fuel and the other 1/3 trying to overcome friction?

Making statements that have no factual proof or in this case, have no bearing on my posting is slightly foolish. Its as if you dont want anyone to think outside the box or succeed. Im not attacking you, just your math. Provide a study on h2 injection in a diesel engine and ill read it with respect.

 

Better re-check your calculator.

I'm not trying to attack the system necessarily, just trying to figure out if there is something I've missed. I looked around and grabbed some general efficiency number that i have seen. At every step of the conversion, you are losing power due to heat, and this method seems to use all of the major power forms. Liquid fuel (chemical energy) > mechanical energy > electrical power > chemical energy (H2 gas), then back to mechanical energy - and loosing energy all the way.

 

Actually his math is correct, yours is wrong KennedyFord.
1/3 x 1/2 x 1/2 is indeed 1/12 - in decimals that would be
.33 x .50 x .50 = .0825 or 8-1/4 percent - which is where he got that number

The numbers he came up with are generally accepted approximate numbers for the efficiency of an internal combusion engine.

Approximately 1/3 of the energy of the fuel burned is lost as heat - out the exhaust and into the metal of the engine block to be carried away by the coolant and dissapated into the air by the radiator. Best case thermal efficiency is around 70% (30% loss).

Approximately 1/3 of the energy of the fuel burned is used to overcome friction losses related to the mechanical components of the engine. This includes everything from the friction drag of the rings on the cylinder walls, to the friction of the lifters on the cam, to the friction of the bearings on the journals, to the friction of the coolant against the water pump blades, to the friction of the cooling can against the air. Again, best case scenario is about 70%-75% efficiency or a 25%-30% loss.

Those two factors compound to give the basic efficiency of an internal combution engine as around 45% (2/3 x 2/3 = 4/9 = 44.9%). Some engines are a bit more efficient than this, but not by a whole lot. Best case is about 55% (.7 x .75 = 52.5%).

His 50% efficiency number for the alternator (amount of energy required to turn the alternator vs. the electrical energy output of the alternator) isn't too far off the mark either. The range is anywhere from 45%-65%. I'm not so sure about the 50% efficiency number he's assuming for the electrolyte cell. It sounds like this is the one unknown factor and that your assertions about the low voltage and ampreage requirements required to produce the H2 are where you disagree with him.

Where his equation breaks down is that he is using the numbers that represent the losses - not the efficiency. Instead of 1/3 x 1/3 x 1/2 - which are the 3 loss factors, the equation should be the product of the 3 efficiencies. The efficiencies are calculated as 100% minus the loss. If the losses are 1/3, 1/3, and 1/2 then the efficiencies would be 2/3, 2/3, and 1/2. Compounding those numbers gives us 2/3 x 2/3 x 1/2 = 4/18 = 22%.

Since the engine is providing the mechanical power to turn the alternator to generate the electrical power to supply the electrolyte cell, then the efficiencies (or losses) do indeed all compound. If we assume BEST case scenario we have .7 (thermal efficiency) x .75 (mechanical efficiency) x .65 (alternator efficiency) = 34% total efficiency for producing the electrical power going into the electrolyte cell.

IF his assumption of 50% efficiency for the electrolyte cell is correct, then multiply that by the 22% (assumed efficiencies) or the 34% (best case efficiency) then we get 11% to 17% total efficiency for the output of the electolyte cell. Better, but still not great. If the home built electrolyte cell is 75% efficient - which I think is probably a real stretch, but we'll assume to be best case - then your overall efficiency is 75% of either 22% (assumed) or 75% of 34% (best case).

That gives us a range of 11% (using average assumed efficiencies) to 25.5% (using absolute best case efficiencies).

 

Thanks CheaperJeeper for saving me some typing. But I will still have to stand by my math. My original quoted numbers of 1/3 x 1/2 x 1/2 = 1/12th were a basic attempt to see just how much energy could possibly go into producing hydrogen.

If 1/3 of a combustion engines fuel is lost as low grade heat and another 1/3 is lost overcoming friction, that would leave approximately 1/3 for motive power [and hydrogen generation]. Then approximately 1/2 of that mechanical power is turned into electrical power in the alternator, and approximately 1/2 of that electrical power is turned into chemical energy (hydrogen) in the electrolytic cell - so 1/3 engine eff x 1/2 alternator eff x 1/2 H2 cell eff. (these are the efficiencies, not the losses - I'm sure the case could be made to gain few percent in one place or another, lose a few percent somewhere else, but this is just a general approximation to see where the power goes)

Even this number is somewhat pointless as there is no way ALL that power would be used for hydrogen. I would be highly impressed if someone drew 80-100 amps continuously from a normal vehicle alternator without it melting down or dying in short order. 100 amps x 12 volts is 1200 watts which is a whopping 1.6 horsepower - pumping that into a 50% efficient electrolytic cell would give you about 0.8hp worth of hydrogen to feed into the engine which would be burned (again putting 30-50% efficiency to the wheels depending on how generous you want to be) So that is a whopping .24 to .4 hp worth of motive power.


The hydrogen electrolysis numbers come from:

http://www.hyweb.de/Knowledge/w-i-energiew-eng3.html

Which quotes: (bold and [ ] is my emphasis)

3.4.1 Present state of the art

- Conventional Water Electrolysis

Process Description : Conventional alkaline electrolysis works with an aqueous alkaline electrolyte. The cathode and anode areas are separated by a micro-porous diaphragm to prevent mixing of the product gasses. Presently in Germany, conventional unpressurised electrolysis utilizes new materials that replace the previously used asbestos diaphragm. With output pressures of 0.2 - 0.5 MPa these processes can reach efficiencies, related to the lower heating value of hydrogen, of around 65%. [this is a well tuned, optimized industrial system] Newly developed diaphragms and membranes from other materials demonstrate, through their good turn off characteristics, relatively good reliability when subject to fluctuating operating conditions. They are therefore applicable in conjunction with renewable energy technologies.

Who offers Electrolysers? Conventional water electrolysers have been in use commercially for many decades. Units with capacities from 1 kWel to 125 MWel are available. The Electrolyser Corporation Ltd. (Canada) and Norsk Hydro Electrolysers AS (Norway) are well established manufacturers of conventional elctrolysers, offering units with very high capacity. Several manufacturers have also established themselves in the 1 - 100 kW range in Europe (e.g. Ammonia Casale, ELWATEC, Hidroenergia VCST (up to 1 MPa), vHS (von-Hoerner-System; up to 3 MPa but also unpressurised).

What is the cost of such electrolysers? Large commercial electrolysers cost between 500-1000 DM/kWel but smaller plants are considerably more expensive. The smallest 1 kWel electrolysers can cost up to 10,000 DM with the price only falling to the 500 DM/kWel figure in the MW range. Operating efficiencies lie in the 50-60% range for the smaller electrolysers and around 65-70% for the larger plants. [again well tuned industrial systems]

It doesn't seem to me that the garage tinkerer could come even close to an industrial system - but even at 50% efficiency, you see how the numbers stack up above.

Heck I remember playing with batteries and electrolysis in science class and even my junior chemistry set - IIRC, it took ~15-20 minutes to fill up a test tube with hydrogen, but I would be surprised if that amount of hydrogen would be enough to fire one cylinder of a large engine for one stroke.

I'm struggling to see where there is any "magic" in this system? On the surface it looks like a huge loss of power. You would be way better off taking the electrical energy from the alternator and using it to drive an electric motor (95% efficient?) to provide direct propulsion. The best bet of all would be to simply minimize electrical load on the alternator, no?

 

I think the whole point of the first post in this discussion was to explore the idea of adding a bit of hydrogen as a supplement to an existing diesel or gasoline burning engine.
The amount of hydrogen needed to supplement the existing air/fuel mixture for modest MPG gains is within acceptable limitations of these "vastly inefficient" electrolysers.
Also, the reference to http://www.hyweb.de/Knowledge/w-i-energiew-eng.html "Present state of the art" is documented on 8/7/1996. In the engineering and scientific arena, that is old news.
The idea of converting your existing egine to burn exclusively hydrogen is a somewhat expensive idea. Try going to www.waterforfuel.com. There is alot of your average Joe garage experiments in the Hydrogen field documented on this site.
Shhhhh. Better not talk too much about all this for fear of the government or worse yet the oil companies will come knocking on your door.

 

Finally..............Good for you friech! Somebody finally got what Troy was talking about !!!

 

Actually I think Corey gets it - I know I do.

I realized from the outset that the H2 & O2 are meant to suppliment (not replace) the petro fuel.

The point we've been trying to make is, if it takes more energy (i.e. burns more diesel and/or puts more load on your engine) to produce the H2 and O2 than what you get OUT of the H2 and O2 when you burn them, then what is the point?

The simple efficiency analysis that Corey and I have attempted to lay out make it pretty clear that the conversion process isn't free, and that more energy goes into electrolyzing the water into H2 and O2 than what you get out of re-combining (burning) them.

The energy has to be coming from somewhere, and the only source to derive it from is your engine. Unless you are going to install another battery to do the electrolysis with - in which case you're still going to have to re-charge it (input more energy into it) sometime, somewhere.

There just ain't no free lunch guys....

 

Yep, there ya go. I seem to be getting a lot of flack over the numbers, but even if you peg them and say the combustion is 80% efficient, alternator is 80% efficient, and H2 cell is 80% efficient, you're still substantially better off by simply not drawing the excess power from the alternator.

You can't escape the three laws of thermodynamics:

Law 1 - You can't win
Law 2 - You can't break even
Law 3 - You can't ever leave the game

PS - by all means, if you want to try it, I'm certainly not trying to stop you, my original question was "is there something I've missed" - besides being +/- a few percent efficiency here or there, it doesn't seem so.

 

Corey YOU are very much correct. I moved my decimal to the wrong location using .3 and it would have been more accurate on .33 and using a calculator instead of my brain.

I just want to try and have a holding bottle with the 2 contacts and plumb the oxygen/ hydrogen directly into the filter assemble for the engine.


1st. I would like to know if I run the terminal on a switch how long 1 gallon of water would last with 13 plus volts running through it and miles till empty.

2. How will the truck react to this plus a full tank of biodiesel

3. Assuming the reaction is less then nitrous characteristics or propane, will I notice any detrimental effects on the truck

4. Can I get an alternative fuel plate and a rebate? lol lol ( I doubt it)

Thank you very much Corey and Cheaper for correcting my ignorance and supporting it with documentation.